Python：随机猜4位密码，1题都猜不出来

问题描述

我是 Python 新手。我有一个程序可以让你猜测一个随机的 4 位密码。例如，如果密码是5530，如果我猜到里面有一个5（例如5111），它说我猜对了2个数字。如果密码是 5535，则表示我猜对了 3 个数字。

但实际上，它将我的答案中的单个 5 解释为对其他 5 是正确的。我希望它只说我猜对了一个数字，如果有两个或三个 5，如果我为 5 输入两个 5530，则两个正确，如果我输入三个 5，则三个正确{1}} 代表 5535。

import random

no1 = random.randint(0,9)
no2 = random.randint(0,9)
no3 = random.randint(0,9)
no4 = random.randint(0,9)

password = str(no1) + str(no2) + str(no3) + str(no4)
count = 0

if no1 % 2 == 0:
    count += 1

if no2 % 2 == 0:
    count += 1

if no3 % 2 == 0:
    count += 1

if no4 % 2 == 0:
    count += 1

print(password)
print(f"Hint: The password consists of {count} even number(s)")

guess = input("Guess the 4 numbers: ")
present = 0

if str(no1) in guess:
    present += 1

if str(no2) in guess:
    present += 1

if str(no3) in guess:
    present += 1

if str(no4) in guess:
    present += 1

if present == 4:

    if guess == password:
        print("Congrats! You have guessed the password correctly.")

    else:
        print("All the numbers are present but not in the correct sequence.")
        print(f"You did not make it! The password is {password}.")

else:
    print(f"{present} number(s) of your guess are present in the password.")
    print(f"You did not make it! The password is {password}.")

PS 我希望我下面的代码可以用 range 和其他东西缩短，但我对此没有信心，所以有人可以帮我解决这个问题吗？

解决方法

改变这个：

if str(no1) in guess:
    present += 1
if str(no2) in guess:
    present += 1
if str(no3) in guess:
    present += 1
if str(no4) in guess:
    present += 1

到：

present += guess[0] == str(no1)
present += guess[1] == str(no2)
present += guess[2] == str(no3)
present += guess[3] == str(no4)

另一种方式：

rand_nums = [no1,no2,no3,no4]
for i in range(4)
    if (int(guess[i]) == rand_nums[i]):
        present += 1

您必须将密码中的每个数字与猜测中的每个数字进行比较。为此，您必须使用索引访问guess。像这样：

if str(no1) in guess[0]:

    present += 1

passwords python